KCET · Maths · Limits
Let the function satisfy the equation \(f(x+y)=f(x) f(y)\) for all \(x, y \in R\), where \(f(0) \neq 0\). If \(f(5)=3\) and \(f^{\prime}(0)=2\), then \(f^{\prime}(5)\) is
- A \(6\)
- B \(0\)
- C \(3\)
- D \(-6\)
Answer & Solution
Correct Answer
(A) \(6\)
Step-by-step Solution
Detailed explanation
\(\because f(x+y)=f(x) \cdot f(y)\)
Put \(x=0, y=5\), we get
\(f(0+5)=f(0) f(5)\)
\(\Rightarrow \quad f(5)[f(0)-1]=0\)
\(\Rightarrow \quad f(0)=1\)
Consider \(f^{\prime}(5)=\lim _{h \rightarrow 0} \frac{f(5+h)-f(5)}{h}\)
\(=\lim _{h \rightarrow 0} \frac{f(5) f(h)-f(5)}{h}\)
\(=\lim _{h \rightarrow 0} \frac{f(5)[f(h)-1]}{h}=f(5) \lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}\)
\(=f(5) f^{\prime}(0)=2 \times 3=6\)
Put \(x=0, y=5\), we get
\(f(0+5)=f(0) f(5)\)
\(\Rightarrow \quad f(5)[f(0)-1]=0\)
\(\Rightarrow \quad f(0)=1\)
Consider \(f^{\prime}(5)=\lim _{h \rightarrow 0} \frac{f(5+h)-f(5)}{h}\)
\(=\lim _{h \rightarrow 0} \frac{f(5) f(h)-f(5)}{h}\)
\(=\lim _{h \rightarrow 0} \frac{f(5)[f(h)-1]}{h}=f(5) \lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}\)
\(=f(5) f^{\prime}(0)=2 \times 3=6\)
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